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First time buyer of Mooney with some questions
I am a private pilot, with 120 hours, working on my Instrument rating, will be a first time buyer. I am interested in getting Mooney M20C for time buildup and cross country. I like them for being efficient and fast. I have few questions.

How much down payment should I have? Most financial institution's websites shows its 15% but when I called one of them today they told me its 20-25% with rate being around 5.5%-6.0%.
How much liquid asset is required for financing?
Can you recommend any banks/financial institutions for financing?
What type of insurance is required by the law if the only purpose of the plane is pleasure under Part 91.
I saw someone selling a share of a Mooney at my local airport but he told me that I can't get insured if I don't have at least 100 hours of retractable gear time. He wasn't too helpful in explaining further. Is that the case with all the insurances or some will insure me with high premiums?

Good questions, although I'm not sure they're a good fit for this site's format (see the [tour] for more information on how the site works). #1-3 and #5 are mostly opinion-based. Or at least, there could be very different answers for different people. The answer to #4 is "none", but there are situations where you might still have to get it, e.g. you rent a hangar and the hangar owner requires you to have insurance. You might get a better response in a forum, especially mooneyspace.com.

For #4, none required by law, but your lender will certainly require some. Answers to #1, #2 and #5 will vary by lender. #3 is opinion-based.

And for #5, it’s not that you can’t get insured with less, but the other partners probably won’t like the rates. My club only requires 10-20 hours of retract time, but the insurance bill is split 80 ways, so it’s not that bad per person.

Ill answer #5 since that is on topic here. The others questions are mostly about personal finance and not aviation.
Insurance companies can require what ever they deem applicable for their own policies. 100 hours is high but not unreasonable, for the average pilot thats about a year worth of flying.
The variable you are missing here is the specific Mooney since there are a lot of M20 variants that will have wildly different premiums. About a year after my PPL I got serious about the possibility of buying a Mooney M20C or E and had no issue getting reasonable insurance quotes with 0 hours of retract time. The quotes were of course contingent on me finishing my complex endorsement.
The issue here is that the owner already has insurance so you are at the mercy of what the policy currently says. The other pilot may also be using it as an excuse to not sell the shares to a low time pilot.
I'd say not much further explanation is needed. Any time the ownership is group based, be it a few people who go in on a plane or a club there are usually insurance minimums to be a "member" of the organization.


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What does "airspace above 14,500 MSL is controlled with few exception" mean?
I was reading Cessna IFR Training program and came across this line that airspace above 14,500 MSL is controlled with few exceptions.

What does it really mean? Class A doesn't start until 18,000 ft, and although Class E is controlled, it is legal to fly in and out of this airspace without talking to anyone, additionally it also doesn't require communication or navigation equipment. So I have following questions.

If I am away from radar coverage area but inside Class E and flying VFR at 15,500 MSL, is that controlled or uncontrolled, if controlled then controlled by who?
Did they mean to say that as you recall class E is controlled?
What is the real world example of airspace above 14,500 ft MSL but uncontrolled?

Class E airspace starts at 14500 feet MSL except for the airspace below 1500 feet AGL or designated and charted to start lower (SETVODA). There are areas like the US Rocky Mountains that are more than 13000 feet in elevation.
Class E is controlled airspace. Sectionals and charts will give you the information for the controlling agency. But just as if not more important, the minimum VFR weather requirements have to match those of controlled airspace (3-152, 5-111).
§71.71 Class E airspace.
Class E Airspace consists of:

(a) The airspace of the United States, including that
airspace overlying the waters within 12 nautical miles of the
coast of the 48 contiguous states and Alaska, extending
upward from 14,500 feet MSL up to, but not including 18,000
feet MSL, and the airspace above FL600, excluding—

(1) The Alaska peninsula west of longitude 160°00'00" W.;
and
(2) The airspace below 1,500 feet above the surface of the earth.

If you are inside Class E airspace, you are in controlled airspace. You would have to refer to your aeronautical charts to determine the controlling agency for specific regions.
They did not mean to say "as you recall class E is controlled."
I cannot find any examples of uncontrolled airspace in the U.S. that is above 14,500' MSL as Class E begins there by default.

Source: FAA PHAK Chapter 15

For #3, you will find plenty of exceptions where the terrain rises above 13,000ft.

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sort: votes


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Can the airplane cabin pressure control be eliminated?
In the future, will it be possible to have airplanes designed from materials that withstand the pressure difference in high altitudes without the need for a cabin pressure control systems?

Could you please give me more details

I think the question is about not having a control which selects 'cabin altitude' (which is typically higher than sea level) but instead always maintains cabin altitude at sea level. Or takeoff level, Which raises its own questions, but I think that's what he means.

@ DJClayworth this is what I meant, but from your answer ymb1 it seems impossible to have such airplane in the future.

Cabins are pressurized to keep the crew and passengers conscious and alive, and not due to any limitations of structural materials.
Without pressurization, everyone on the airplane would be unconscious from lack of oxygen.
Even if the plane were sealed at the pressure of the departure airport, you still need a pressure control system to continually bring in fresh air, pressurize it, and deliver it to passengers.   There just is not enough air in a sealed airplane to sustain all 150+ people for a multi-hour flight.  Pressurized, fresh air is an absolute requirement
So I don't see how you're going to do away with a pressure control system, as your question seems to suggest.

from your answer it seems impossible to have such airplane in the future ... thanks anyway

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How does a turbofan with afterburner continue to function in extreme humidity and rain?
From what I understand, the fan(s) collect(s) and compress(es) air into the compression chambers, where fuel is added to the air & combusted to create an even bigger pressure/thrust, right before the fuel-air mix hits the final chamber where even more fuel can be added & combusted (if you go +afterburner).
My question is this: If it's very wet out, and the engine is sucking up multiple gallons of water per second, then exactly how is combustion achieved with all that water contaminating the air?

A turboFAN engine with an afterburner seems like an unusual combination. Do you have an example of an engine that is built that way?

@RalphJ https://aviation.stackexchange.com/q/36903/4108

The air is compressed in the compressor section of the engine and heats up. Hot air is able to hold disproportionally more water in gaseous form than cold air, so all that water will vaporize and enter the combustion process together with the air. Since it is already vaporized, it will not absorb that evaporation enthalpy again that makes water such a good extinguishing agent. On the other hand, since the evaporation enthalpy was sucked from the air in the compression process, the air will be cooler when it enters the combustion chamber, so more fuel can be burnt (up to the stoichiometric ratio of course) and power can be increased.
This trick was used on high performance piston engines and early jets. The Boeing 727 would inject water into the compressor on purpose on hot and high take-offs to increase engine thrust.
If the amount of water entering the engine overwhelms the absorbing capacity of the hot air, the engine runs the risk of flaming out. Regular precipitation does not pose this threat, however.

"Regular precipitation does not pose this threat, however." Relevant video

How about turbofans with a larger bypass ratio? Do they rely on premixing, or centrifugal water removal, or?

Fun fact, water injection does roughly the same thing as nitrous (NOS) in a high-performance car

@Sanchises: The centrifugal water removal comes standard on turbofans. That is one reason why on the 727 the water was injected after the first compressor stages. The other is that freezing of the water inside the intake could be avoided this way.

Is the water vaporization increasing the volume and thus the pressure a factor here?

@rotard: water injection in a jet works pretty much like water injection in a gasoline engine (which is also done at times). In both cases, it cools the incoming air stream, so when it gets used for combustion, it's denser. NOS, by contrast, splits into nitrogen and free oxygen, which increases the percentage of oxygen in the air.

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Why are the wings of some planes changing width?
I was travelling recently in a Boeing 737 and I noticed one thing that I didn't understand. After the takeoff the wing flaps started retracting into/under the wing, making it narrower. The opposite happened before the landing.
You can see what I mean in this video:

The question is:
Why are the flaps hiding into/under the wing?
I understand that the bigger wing area is better during takeoff and landing. But why is the wings' size reduced during the cruise? Shouldn't bigger wings provide better lift during the entire flight?
Some clarification after comments:
I'm asking specifically why are the flaps retracted and not just remain parallel to the wing

Possible duplicate of Why and when to use flaps?

@ymb1 Judging from the last paragraph, it sounds like he's just asking why the flaps are retracted for cruise, which seems like a duplicate of the other question.

Yes, @ymb1 is right. I knew that flaps in the position not parallel to the wing would increase drag. However, I didn't realise that bigger wing itself results in more drag too.

@pajonk So, you're asking why they're retracted rather than just raised to be parallel with the wing? Some flaps do exactly that, actually, just hinging on a fixed point instead of retracting forward underneath the wing. That's pretty common on light aircraft. That kind is called plain flaps, while the kind you see more commonly on airliners are called Fowler flaps (or some variant thereof, at least.)

@reirab Yes, that's right. I'll try to clarify that in the question.

@pajonk Ok, I've retracted my duplicate vote.

There's a simple rule to remember.  Lift is not free.  You have to pay drag to get some.  The more you have, the more it costs you.

Well, yes, bigger wings to provide better lift, but the also produce more induced drag in the process.
The wings on an airliner are optimized for cruise in high subsonic and transonic flight where a slender, swept wing works well.  While this is great for cruise flight, the trade-off is this style of wing requires a very high approach speed for landings which in turn require very long runways to accelerate the airplane on to reach rotation speed for takeoff or to decelerate the aircraft on once it has landed.
The Boeing Company successfully addressed these problems in the early 1960 with the development of the 727 airplane as a regional airliner.  It made use of a type of flaps called Fowler flaps (see Fig 1) in concert with leading edge extensions.  Fowler flaps. These style of flaps consist of a series of segments attached to tracks or support linkages running chordwise, allowing the flaps segments to extend and retract by rolling along said tracks.

Fig 1. Typical Fowler flap installation
When deployed these give the effect of changing the airfoil shape from a slender, slightly cambered airfoil into a wide airfoil with a large camber.  Fowler flaps have an additional advantage to them in that partial deployment creates a large increase in lift with limited additional drag, very useful for takeoff, while when fully deployed they create a lot of drag in addition to higher lift.

but not to the extent which the 727 did.  The reduced approach speeds and ability to get into short airfields from that was one of its chief selling point.

Induced drag depends on the lift and speed and the lift is always the same as in level flight it needs to exactly balance the weight. So leaving the flaps extended would not increase the induced drag. It would increase the parasite drag because the camber would be too large for the amount of lift needed and the wetted area would be larger.

Flaps are a way of changing the shape of the wing so that it is able to provide more lift at lower speeds and higher angles of attack. This is important both on take-off and, especially, landing where the aircraft is moving relatively slowly.
The trade off is that flaps dramatically increase the drag.
Bear in mind, that commercial airliners spend most of their flight time cruising at, more or less, constant speed and altitude. At cruise, there is no value in 'more' lift, they need exactly enough lift to support their weight at an efficient cruising speed. So, the wing shape is designed to do this with the minimum possible drag.
Flaps increase the lift at low speed so the aircraft still has enough lift  to support its own weight when approaching for landing without needing to be traveling at a speed which would make safe landing more difficult and require a very long runway to slow down after touchdown.

The "flaps up" configuration represents the wing's base shape. This is the shape designed to produce the intended flight characteristics including required lift at minimum drag under the aircraft's designed cruising flight conditions (airspeed/altitude).
Take-off and landing is best performed at the lowest speeds which are practically attainable (take-off and landing distances increase dramatically and non-linearly with speed). The "flaps down" configurations (note the plural) increase the effective wing area, but also change its profile. Extending flaps increase lift but also increase drag. Most aircraft have multiple flap settings, providing a progression between "clean" (flaps up) and full flaps. Full flaps are only used for landing where high drag is desirable because it helps slow the plane down after touchdown. Intermediate settings are used for take-off because they aid lift with a minimal drag penalty and are used in the early stages of approach as the aircraft is slowing down.
Flaps aren't simply enlarging the wing. Most flap designs leave gaps when fully or mostly extended. These gaps allow some air from under the wing to flow up through the gap and join the air moving over the top of the wing. This helps the airflow remain "attached" to the top surface of the flap. Without this feature, air on the top surface can "detach", become turbulent, and no longer aid lift.

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Examples of Suzerainty
Under Article 2 of the ICAO (chicago) convention, the Territory of a State is considered to be those areas under which the state has suzerainty (control over the foreign policy and international affairs).
Are there any current examples of this in practice?

That is a legal term, not an aviation term.  It simply happens to appear in an aviation document.

I agree. This asks for examples of a political/international construct, not for anything aviation-related. Also, the wikipedia article already linked gives some historical examples.

question updated (current examples).  I'm interested from an aviation perspective (as it determines which jurisdiction is applicable when flying through different airspaces)

Very interesting use of the word "adjacent".  Adjacent generally means "next to" rather than "part of".  Legal clarification  may be in order.  As far as jurisdiction, this would clearly be defined by boundaries, with the exception of disputed areas.  I would certainly do homework before going there, and try to be aware of the potential dangers.  There may also be a "pecking order", such as ground control modifying flight plans as needed.  It would depend a lot on what "different airspaces" you had in mind.

I don't really understand what you're asking, but you might get a better response on law.SE

Good examples are Monaco, under France suzerainty, or San-Marino (maybe Vatican) under Italy one's. There are lots of them.

The ICAO is the International Civil Aviation Organization founded in 1947.  The statement "for the purposes of this convention the territory of a State shall be deemed to be the land areas and territorial waters adjacent thereto under the sovereignty, suzerainty, protection, or mandate of such State" has applications to current world issues.
Though it is understandable that some may feel this is unanswerable as a "aviation question", it is relevant and sparks interest.
Rules are made by people for people.  This wording, written over 70 years ago, may be subject to review.  Key words are "adjacent" and "sovereignty, suzerainty, protection, OR
mandate".  As with any Standard Operating Procedure, an improved version may supersede it.
If an area is in dispute, that does NOT mean rules for air space over it can not established and abided by, especially for civilian air craft. Cooperation of the parties involved is a possibility.


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Is the the cost of taking a passenger jet out of long-term storage for two years significantly more than a C check?
There are separate procedures for "parking" (short-term storage) and "storage" (long-term storage), as Airbus calls them; this question is about the latter. Assume that close to optimal storage conditions (as specified in the procedures) are used, as would be typical of a commercial airline putting a passenger jet into long-term storage.
According to this article it takes about 200 hours to put a typical passenger jet into long-term storage. Though maintenance is still required, presumably the maintenance costs during storage will be considerably less than if the aircraft were being flown every day.
According to this comment on YouTube, taking an aircraft out of long-term storage is very expensive; considerably more than the biennial C-check that takes up to 6000 man-hours:

It is a lot more, in fact the longer the storage time the more expensive it gets.  Eventually it becomes cheaper to buy a new plane of the same type from the manufacturer, though that point takes fifteen or twenty years to reach.

Is this true, assuming that the jet has been stored (and maintained in storage) per the manufacturer's recommended procedures? If so, why is it so expensive? When taking an aircraft out of long-term storage and restoring it to flight condition, what additional procedures are needed that would not be part of a C-check had the aircraft stayed in service? Are there any C-check procedures that could be delayed (because the aircraft was not in use) to offset this cost?
If the return-to-service versus C-check cost ratio varies dramatically amongst passenger jets, I'd like to hear about examples with the highest ratio you know of.

What authority does the "comment on YouTube" have?

@RalphJ I am not aware of what level of knowledge that commenter has. He sounds very sure of himself, which initially inclined me to believe him, but on further thought I'm starting to have my doubts. Thus my question here.

This question can only be answered with opinions because it depends on the aircraft, it's age, how long as it been in storage, etc. There isn't one answer for this. Voting to close.

@JuanJimenez Ok, it no longer depends on how long it's been in storage and I'm now asking not about the cost per se, but the ratio between the return-to-service cost and the C-check cost, which also will vary in the same way with the aircraft age/type/etc.  (This was my original intent in the first place.) If you can think of other ways to improve this question to make it more easily answerable, please suggest them.

Suspect not an answer for question as written but this book https://www.examiner.com.au/story/3127169/aviation-career-takes-off-in-print/ includes a multi month struggle to get a small commuter turbo prop airworthy again. It seems in that case due storage conditions they had to strip and replace every rubber seal and hose in the entire airframe despite nominally dry conditions, complicated by the remote location from both spares and engineering support.

@cjs The question is still opinion-based. Each aircraft is different. Storage period, location (temperature, humidity, etc.) and how it was preserved makes a huge difference. You need to significantly narrow it down if you want an answer. The way it is I still think it should be closed.

@JuanJimenez I referenced detailed vendor procedures for putting aircraft into long-term storage; these cover environment. I think you're really starting to reach here with your assumption that the airline wouldn't follow these procedures unless I specifically said so. You are clear that this is not about random aircraft but about expensive passenger jets like A320s and A350s, right? At any rate, keep giving me all the things you think I need to specify to make it less "opinion-based" and I will keep adding them to the question.

@cjs I stated no assumptions, period. Pick ONE aircraft (make, model and year), ONE storage location and ONE storage period as an example. That would narrow it down to produce a specific answer.

@JuanJimenez I'm not looking for one specific answer here for, e.g., a Delta 737MAX manufactured in 2018 and put into storage in 2019; I'm just trying to get a general sense of, "is it really massively more expensive to bring a jet out of storage after a couple of years than to keep it running for a couple of years." You seem to be trying to get me to post dozens of very similar overly-specific questions in the hope that some could get answered.

My vote to close stands. I'm out of this.

As a partial answer to 'why is it so expensive' part, based on reading including this and personal experience bringing ground based equipment out of storage.
If everything has been maintained IAW the manufacturers documented processes while parked then returning it to flight things could be straightforward, but parked aircraft are normally not flying because the money ran out and then things get complicated:

Maintenance was not in fact in date when parked
It was stored because something expensive broke
While stored parts were removed/sold/stolen
Preparation for storage was done by someone who knew they might not get paid (see stolen/swapped parts above)
May have been parked in the lowest cost location (road salt, paint overspray, forklifts or trucks bumping it, or just plausible risk of these)
The aircraft was stored because it was worn out/unreliable/cursed.

These mean that a number of hidden risks may exist even if a look at the calendar says it would be fine to service and fly. Even if stored correctly:

A parked aircraft will not get freeze dried at high altitude regularly, and tanks/voids will pull moisture in during day/night cycles resulting in corrosion, mold and even live plants. Some fluids can also decay or separate left static for too long.
Most connectors are weather resistant when fitted, but removed equipment can leave plugs to corrode if not sealed correctly.
Dirt and contamination in fluid and air systems can build up, rather than being regularly flushed into filters during use. They then sit until at some future point they break loose as a solid chunk blocking part of a system. Blockage may blow back out during shutdown so replacing the blocked part may not get rid of it.
Rubber weather seals decay for a combination of leaks and glue like residue sticking things in place.
In hydraulic systems seals and hoses often change if left to dry out, potentially producing cascades of leaks bringing system back to operation. Leaks may cause further damage to surrounding systems.
Bearings and valves left to sit under load can end up stuck due combination of settling and oil/grease displacement for erratic operation.

This means bringing an aircraft back into service can quickly escalate into opening many normally sealed areas of the aircraft to check condition and repair/replace hardware and structure. This work will cause damage to corrosion protection that then needs to be repaired and the debris from that needs to be flushed out. If poorly planned this may need to happen several times.
If the repairs become extensive then original serviceable but imperfect fixtures/covers etc may need replacing/repair because they are now obvious to those assessing the work.
All of this means that costs of a return to service can be arbitrarily high, as many vintage car owners can attest.

Thanks. This is helpful, though I was envisioning a situation where a passenger jet was properly stored and maintained during storage per the manufacturer's recommendations, as I'd imagine any major airline or leasing company would do with an aircraft that they're intending to bring back into service rather than dispose of.


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Does climbing or descending affect the rate of turn?
In the past, I used to have this idea that if all other factors remain constant, then climbing increases the rate of turn while descending decreases it. This idea was most likely planted into my mind from my past flight simulator experience.

Assume that the bank angle remains constant, so does the speed (TAS), and the aircraft is in a steady climb/descent (constant vertical speed). Will the rate of turn be any different from what it would be in level flight?
It is known that lift is less than the weight of the aircraft in a climb/descent, since thrust/drag are bearing the remaining weight. It is also known that the horizontal component of lift is what provides the centripetal force for a turn to take place. Does this have any affect on the rate of turn? what other factors affect it? or is the rate of turn unaffected?

Your first sentence (for a student pilot) indicates that ROT increases in a climb because airspeed is dropping.  Very important to keep an eye on airspeed.  Realisticly, we can start with climbs and descents of less than 10 degrees.  Cos 10 is 0.985 so lift requirement does not change much but sin 10 is 0.173.  This is the percentage of your aircraft weight as gravity "thrust" in a dive and what the engine must lift in a climb.  Climbing or descending turns can be done by adding or subtracting power.  Try a few in the simulator,  let us know how you do.

That's most likely how that idea got into my mind. rate of turn depends on TAS; in my simulator (X-plane), in climb after takeoff, I have to use relatively small bank angles for rate 1 turn, because my TAS is low. When I'm descending, I'm perhaps at 20~30k feet, and so the TAS is much higher. So for the same bank angle, I get unacceptably low rates of turn. I understood this concept some time ago while studying flight mechanics, but that's when I had this new problem: What if the TAS remained constant? Will the rate still vary in climb/descent, compared to level flight?

This new problem that I had could not be answered through simulator - even if it's physics was perfectly identical to real life. For any appreciable changes in cos ∅, you need extremely large values of ∅ (angle of climb/descent), and at those extreme values, your TAS is changing all the time; it's practically impossible to maintain a constant TAS. So looks like this question cannot be answered by experimenting in simulator.

Thanks for the recent additions! I am currently in the process of analysing the new answers and chat discussions.

Bear in mind that you can’t keep all other factors constant, eg either airspeed, rate of turn and turn radius can’t all stay the same if you climb or descend.

This is one of those situations where it is helpful to extrapolate to the most extreme possible cases.
Imagine an aircraft in a descending turn, with the descent path incrementally getting steeper and steeper.  I.e. the glide ratio is incrementally getting poorer and poorer.  The flight path is incrementally getting closer and closer to a vertical rolling dive, with the direction of the roll being toward the lower wingtip.
In less extreme cases, when the flight path is below horizontal, but not yet vertical, note that a non-zero roll rate is needed just to maintain a constant bank angle, with the direction of roll being toward the low wingtip.  If the roll rate decreases below the required value, the bank angle will decrease.
At the extreme limit case where the flight path truly is aimed vertically downward, the wing is generating no lift, and the aircraft's weight is entirely supported by the drag vector.  In this case turn radius is zero, and the bank angle is undefined, and the roll rate is no longer constrained by the dynamics of the turn.  The pilot can set the ailerons to any position he or she desires, and the roll rate will be limited only by the aerodynamic "damping" effect in the roll axis.
Note that if the airspeed is stipulated to be constant, then as the flight path gets progressively more vertical, the component of the airspeed vector which is tangential to the circle of the turn-- i.e. the horizontal component of the airspeed vector-- gets progressively smaller.  This means that a progressively smaller centripetal force component is needed to obtain a given turn radius, and also that a given centripetal force component will drive a progressively tighter (smaller) turn radius.
Of course the situation is complicated by the fact that the centripetal force component is not constant, because as the flight path gets progressively steeper, a progressively larger portion of the aircraft weight is borne by the drag vector rather than lift vector, so the lift vector gets smaller.  Still, the net effect is that due to the decreasing turn radius, the turn rate will increase as the flight path gets steeper, given the assumption of constant airspeed and constant bank angle.
Essentially all the same logic applies to a climbing turn as well.  The main differences are that 1) the climbing turn, it is the thrust vector rather than the drag vector that supports some of the aircraft weight, and 2) in the climbing turn, the direction of roll required to maintain a constant bank angle is toward the high wing tip, not the low wingtip, so if the roll rate falls below the required value, the bank angle will increase, and 3) the extreme limit case is a vertical rolling climb with the direction of roll being toward the wingtip that was the higher wingtip when the flight path was less vertical.  The end result will be the same-- the turn rate will be higher when the flight path is more vertical than when the flight path is entirely horizontal, given the assumptions of a constant airspeed and a constant bank angle.
In the real world, of course, the idea of using one given constant airspeed for a wide variety of trajectories ranging from vertical rolling climbs to ordinary horizontal turns to vertical rolling dives doesn't make a lot of sense.

Indeed considering the extreme cases is how I understand flight mechanics; I couldn't wrap my head around the extreme cases in this particular situation, and your answer really helped me with that. However, I have a question. You state that the net effect of the two factors (reduced forward component of velocity and reduced horizontal component of lift) favours the former, causing the turn radius to decrease. I couldn't agree with that statement, since I believe that both factors must change in equal proportion, cancelling the effects of each other. (Continued...)

Let ∅ denote the angle of climb or descent. We know that the forward component of velocity is a function of cos ∅. We also know that lift (and therefore it's horizontal component) is also a function of cos ∅. So we know that these two factors change proportionally. Then how does one outweigh the other? Thank you for reading.

@AdityaSharma -- the answer admittedly isn't fully complete yet. It may take me quite some time to post a revised version

No issues! just wanted to adress that I messed up somewhere: The forward velocity (forward TAS) is a function of cos ∅, but the radius of turn is proportional to the SQUARE of forward velocity; I completely forgot about that.

@AdityaSharma -- Re comment above "the answer admittedly isn't fully complete yet"-- since a new answer has now been posted that reaches a different conclusion, it may be time for that long-promised revision of this answer--

@ quiet flyer - I have deleted that post. I thought I had a good answer but my equation did not reflect the fact that, as you have noted, the both lift and turn rate go to zero as the pitch angle approaches vertical.

"At the extreme limit case where the flight path truly is aimed vertically downward, the wing is generating no lift." This is false.

@MarkJonesJr. -- why?

Some discussion relevant to this answer is taking place in the related chat room for another related answer https://chat.stackexchange.com/rooms/152051/discussion-between-phil-crowther-and-quiet-flyer, starting especially here https://chat.stackexchange.com/transcript/message/65346615#65346615

(I'm really not entirely sure that this present answer is correct, see the chat for more, and stay tuned for a future update!)

Lift occurs perpendicular to relative wind.  In the case of flight path aimed directly downward, there is air flow over the wing, so lift can be created.  This is precisely how an airplane can pull out of a loop, by generating enough lift to move the nose from the vertical to level flight.

@MarkJonesJr -- the point is that we are considering steady-state situations, not dynamically changing ones.  In the steady-state case of a vertical descent, the wing must be at the zero-lift angle-of-attack, otherwise we would be generating lift which would cause the flight path to curve away from the desired vertical line, just as if we were pulling out of a loop.  Unless we want to get off into the weeds with additional trivia like maybe the thrust line isn't parallel to the relative wind at the zero-lift aoa etc but that's not the main point at hand here.

That's still false. An airplane that is rolling is creating lift.

@MarkJonesJr. -to roll, you have to overcome the resistance from aerodynamic damping. Requires deflected ailerons, down on the rising wing and up on the descndng wing. Once roll rate is constant rather than increasing, net roll torque (including roll torque from aerodynamic damping) must be zero. If going straight down, net aerodynamic lift force is also zero. Note also that when going straight down, if you are causing the roll rate to increase, net roll torque cannot be zero, but net aerodynamic lift force is still zero.  + and - lift forces from the two wings cancel each other out.

(And all the same for going straight up, naturally.)

For several years, I have been working on a single-vector version of a Flight Simulator Module which uses speed as the primary vector where Thrust, Drag and Gravity_x_Sin(Pitch) change the magnitude of the Speed Vector and Lift and the Gravity_x_Cos(Pitch) change the direction of the Speed Vector. As seen in this example, Lift acts along the Pitch Plane and Gravity_x_Cos(Pitch) acts along the Vertical Plane.
The OP asked a question which I had also been wondering about. I wanted to both validate my Flight Simulator Module and to compute the Lift settings necessary to hold the Airplane stable in a climbing turn.  My initial goal was to simply to mention a few things and hope someone else would come up with a definitive answer.
However, while doing so, I felt that I had stumbled across a vector-based solution when I realized that the horizontal rate of turn is the result of two factors - the horizontal component of the Speed Vector and the horizontal component of the Lift Deflector. In a level banked turn, the Total Lift required equals Gravity/COS(Bank). The vertical component of the Lift equals the Gravity Deflector and the horizonal Lift equals Lift_x_SIN(Bank). Thus the Horizontal Turn Rate equals the (Horizontal Lift/Speed)x(PI/180). The entire equation is:

Turn Rate = ((Gravity/COS(Bank))_x_SIN(Bank)/Speed)x(PI/180)

This exactly matches the result given by the "old standard" equation of:

Turn Rate = TAN(Bank)x(Gravity/Speed)x(PI/180)

When Pitch changes, two things happen: (1) the Lift required to maintain climbing Flight decreases since the Gravity Deflector is Gravity_x_Cos(Pitch); and (2) the Horizontal Component of the Speed Vector also decreases and equals Speed Vector_x_Cos(Pitch).  Since Cos(Pitch) appears in both the numerator and the denominator, my conclusion was that the Turn Rate stays the same.  While I initially did not like this answer, because I assumed that the Turn Rate should decrease to zero, I later realized that it is the Turn Radius that is decreasing to zero.  (And if Turn Rate did decrease, Turn Radius might remain the same or even increase - an equally unappealing result.)
So I set about to create an example of my flight model which would help determine the correct answer.  As can be seen, the latest example appears to confirm my initial conclusion - that the Turn Rate remains identical and that Turn Radius decreases.
This appears to be the answer reached by other respondents - apparently confirmed with some "heroic" real life tests.
I had been hoping to further validate my conclusions with a mathematical explanation of what the Flight Simulation Model is doing (it uses linked rotating objects) - but have not done so yet.  In the meantime, I have been able to use this knowledge to compute "autopilot" values for the coefficient of Lift which keep the airplane in a steady climb - which seems to be a further validation. So, at the moment, I am happy with what I have learned and hope that it is useful to others.
Note that this is largely a hypothetical exercise with very little real world application since most aircraft will be unable to maintain a fixed speed at Pitch +/- 15 degrees and because air density almost always decreases as you climb and increases as you descend, both of which affect your turn rate.
Thanks to everyone (especially quiet flyer) for their comments and suggestions!

Related questions that address spherical geometry of pitching while banked, etc https://aviation.stackexchange.com/q/97987/34686, https://aviation.stackexchange.com/q/98210/34686, https://aviation.stackexchange.com/q/16531/34686, https://aviation.stackexchange.com/q/66945/34686

It would be super-cool if you modified your demo programs to include readouts of bank angle, heading (maybe that's there already), and also cumulative degrees of rotation of pitch, yaw, and roll since start of simulation.  Would allow hands-on exploration of some of the related questions linked in my earlier comments.

Let us continue this discussion in chat.

Some interesting comments have been moved to chat https://chat.stackexchange.com/rooms/152051/discussion-between-phil-crowther-and-quiet-flyer and the conversation will likely continue further there --

Does climbing or descending affect rate of turn

Many people new to flying will only pull the yoke to climb and push to descend, resulting in an decrease in indicated airspeed while climbing and an increase while descending.
Since rate of turn is proportional to bank angle and inversely proportional to speed, what is going on here?  With higher AoA, the same centripetal force is created for a lower airspeed, resulting in a higher rate of turn.  Opposite for lower AoA.  All fine and good, as long as you do not stall in your climbing turn, or exceed V manuvering in your descending turn.
Fast forward to the airliner simulator, 30,000 feet.  "Unacceptably slow rate of turn".  At higher altitudes, air density Rho comes into play in a much more significant way.  TAS becomes increasingly greater than IAS as density decreases.  If IAS is kept constant, TAS increases and rate of turn decreases with altitude, and while descending, TAS decreases and rate of turn increases (within the constraints of less than +/- 10 degrees from the horizon).
However, if TAS is kept constant, we can model our climbing/descending turn as a cylinder rather than a circle.  Flying at the same TAS, more distance must be covered if the cylinder radius remains the same as the horizontal circle.
What follows is an amazing group of cosines canceling each other out:

Cosine angle of climb/descent × Weight = Lift component
Lift is less than weight in a climb/descent.
less centripetal force

Cosine angle of climb/descent × TAS = horizontal TAS
 if the radius of the horizontal circle component of the cylinder remains the same, horizontal TAS is slower
 rate of turn (around the horizontal circle) is lower

But what of the rolling component as pitch to the horizon increases?  In a practical application of climbing in the range +/- 10 degrees, it may not factor in significantly$^1$.
$^1$ this would be revisited if radius of turn decreased

I think I didn't make my point clear. I know that the rate of turn varies with TAS, and so I'm interested in the case of a constant TAS. Will the rate of turn still vary in a climb/descent, compared to in level flight?

@AdityaSharma some ROT reading as for how airliners turn.  They seem to limit bank angle. to 25 degrees, and just take more time turning at higher TAS.  Good to work with you.  I learned some as well.

Thank you for that link, that's some interesting information!

Just noticed something in your second point of the cosine section. Wouldn't the radius of the cylinder reduce, if the forward component of TAS is made to reduce, while keeping the same net TAS? quiet flyer pointed this out, and I found it reasonable (for instance, as the flight profile tends to vertical climb/descent, the turn radius tends to zero).

@AdityaSharma originally thought this too.  But horizontal component of centripetal force reduces as well (because wing lift is reduced), both by cosine climb angle, so radius should not change.  This would be a different case from a spiral dive, where the pilot tries to maintain altitude with increasing amount of elevator.  If bank angle and TAS are constant, should not radius be stable?

Bank angle is constant, TAS is constant, but the forward component of TAS is not - it has reduced. The centripetal force has also reduced, but then you also don't need as much centripetal force with a reduced radius - for the same rate of turn. Imagine an aircraft turning in an 89° dive. The centripetal force is now much less, but if the radius has reduced, for the same rate of turn - the centrifugal force required has also reduced. So the idea of a reduced cylinder radius made sense to me.

I believe that the turn radius depends upon the bank angle and the forward component of TAS, not the net TAS,. Although, now that I think about what you said, it sounds very reasonable, and I might be wrong (also, intended to say centripetal force in the previous comment, not centrifugal).

@AdityaSharma well, with all the progress, there is certainly no right or wrong here.  At +/- 10 degrees, cosine values are little changed, so ROT should be only slightly changed.  One for the test pilots to confirm.  Thanks.  Robert.

Sure thing. Now this is nothing to go by, but just wanted to let you know, I did the following in the simulator: In a boeing 737, I climbed to 20k feet, deployed the thrust reversers, landing gear and speedbrakes. Next, I started an 80° pitch down dive, while trying to maintain a bank angle of 20-30°. The peak speed exceeded 400 KTAS, but for the few seconds for which I could stably sustain this manoeuvre, I got a rate of turn of around 2° per second, which I found to be really interesting.

@AdityaSharma excuse me, just needed to check some numbers.  Try ROT 30,000 feet 30 degrees 325 KTAS level circle ROT, then engine idle, 325 KTAS descending circle.  I think you will find, as IAS increases, AoA must decrease to hold 325 KTAS.  See how many turns you can do down to 10000 feet.

Too late, already tried 250 KTAS (barely avoided stall). Here are the numbers: 2.69°/sec in level flight, 2.79°/sec in idle descent. Additional details if you're interested: Used autopilot throughout for it's superior precision. Maintained 245 - 255 KTAS in both cases. The numbers are corrected steady state values, accounting for all possible sources of errors known to me. Manoeuvre used: 180° turn with stable 30° bank angle (didn't include times when the plane was banking and unbanking). The AoA will definitely decrease as you said, since the dynamic pressure increases.

@AdityaSharma    Just occurred to me:  the plane in the level turn is actually pitched up a bit.  Ha! What was your AoA at altitude?  I'm sure all at Aviation Stack Exchange will very much appreciate data from your work.

Let us continue this discussion in chat.

climbing increases the rate of turn while descending decreases it

Vertical speed affects the rolling moment equilibrium as explained here. Climbing will roll the airplane into the turn and descending will roll it out of the turn. However, the changing bank angle should be corrected by the pilot, just as is the sideslip angle which might develop when turning.
However, the opposite is true with regards to the maximum rate of turn!
Turning requires extra lift to add centripetal force, which in turn increases drag and requires extra thrust. Climbing also requires more thrust while descending frees up potential energy which would accelerate the airplane if the engine weren't throttled back and allows to create more lift with full thrust for a tighter turn. Unless your angular velocity is limited by lift and excess thrust remains, the more thrust goes into climbing, the less will be available for turning.

Maximum turn rate diagram for two flap settings: 0° (blue lines) and +10° (red lines). Own work.
Here you see the computed turn rate (Drehrate) over Mach for two flap settings (red and blue lines) and different climb rsp. descent speeds. While limited thrust at 10 m/s climb speed does not leave enough excess thrust to pull even 2 gs (with a turn rate below 4°/s), with -40 m/s climb speed (= 40 m/s sink speed) so much more thrust is freed up that the maximum load factor of 2.66 is easily exceeded and the turn rate reaches 7°/s. I did this calculation for a HALE platform which explains the low g limit. Values above Mach 0.75 are not realistic because I did not properly include transsonic effects.

This is interesting info but does it really address original question which specified "Assume that the bank angle remains constant, so does the speed (TAS)"?  I don't see how the graphic shows how for a given bank angle and TAS, the turn rate will vary according to whether the aircraft is descending, climbing, or flying level.

Likewise the arguments you give in the text of the anwer seem focused on total G-load (which is related mainly to bank angle, but also to climb or descent angle) that can be achieved with a given amount of thrust before the aircraft is forced to decelerate due to excess drag.  But what if we assume unlimited power is available and want to explore how turn rate varies with climb or descent angle, for a given TAS and bank angle?

@quietflyer Acknowledged. Please look at the first paragraph now, I added it to account for unlimited thrust.

Perhaps our biggest challenge is that we are having to make a lot of unrealistic assumptions to answer this question.  First, unless you are flying a jet fighter, you will be unable to make a sustained speed climbing turn at more than, say, 15 degrees of upward Pitch and you will be climbing into thinner air.  Conversely, unless you have dive flaps, you will be unable to make a sustained speed descending turn at more than, say, 15 degrees of downward Pitch.  So we are working within a very limited range. To answer this question, we have to remove all realistic elements from our flight models.

@PeterKämpf  First of all, thank you for taking your valuable time to write an answer! the information you shared is very useful from a practical standpoint. However, it doesn't seem to answer my original question (even after the edit). In particular, it doesn't explain the effect of climb/descent on rate of turn assuming a constant bank angle, VS and TAS,  as quiet flyer mentioned above.

@AdityaSharma Since you need less lift in both climbing and descending flight, less sideforce is available from the wing for turning, assuming a constant bank angle. Now think of a spiral: The distance the airplane has to fly for a full turn becomes longer once it climbs or descends, so with the same TAS the time for a full turn also becomes longer, assuming the same turn radius. Both effects conspire to reduce turning rate. Would this answer your question?


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If an aerofoil is placed in a narrow wind tunnel, how will that affect the lift it produces?
This is NACA 0000 aerofoil:

This is just a straight line - roughly what a paper plane's aerofoil section looks like (I chose this aerofoil for its simplicity). If we put this aerofoil in a wind tunnel, then this is my approximation of what the air-flow around it might look like:
(Infinite wind tunnel - ignore ground effect)

$$(AoA=12°)$$
The air-flow isn't ideal; there is flow seperation already at the leading edge. But still, the aerofoil will produce a net positive lift.
But what happens if we put this aerofoil in a narrow wind tunnel?

The pressure distribution appears to have been reversed: high pressure above and low pressure below.
This is explained as follows: through a convergent duct, the air-flow velocity (and thus dynamic pressure) increases. This is accompanied by a reduction in static pressure, such that the total pressure remains constant (the opposite is true for a divergent duct).
Is this true? and so will the aerofoil produce a downforce instead of lift in the narrow wind tunnel?

This is a case where working backwards from actual wind tunnel data may be more helpful.
A higher pressure stagnation zone will form underneath the plate.  This is the ground effect, increasing lift.  Air velocity will increase at the exit of this area (at the trailing edge).
A lower pressure area forms above the plate, with a lower (or even reverse) air flow at its exit.  This will increase lift.
The Formula 1 car is producing downforce (or a reverse ground effect).
So, Bernoulli can't explain everything.  Local pressure may be a more accurate way to predict lifting effects.
However, if a more smoothly cambered airfoil is chosen, interference with upper airflow may indeed reduce lift, especially at higher Reynolds numbers.

Read on about biplane wing interference effects.

"A higher pressure stagnation zone will form underneath the plate." - but the pressure never increases at the entrance of a Venturi, it only decreases, there is no stagnation (based on my limited knowledge of the Venturi effect). What would make this setup different from a Venturi?

@AdityaSharma it may be that the trailing edge is where velocity increases, and pressure decreases.  The upper wall may indeed reduce lift (even with the flat plate).  I think the upper wall may cause a stall at a lower AoA.

"It may be that the trailing edge is where velocity increases, and pressure decreases." - In my opinion, that shouldn't be the case. To justify my opinion, I would first like to summarise why the velocity increases and pressure decreases in a Venturi. There are two principles in action: 1) Law of conservation of mass, and 2) Bernoulli's principle. Law 1 tells us that mass flow rate (of an ideal fluid) through a streamtube will always be equal at every section of the tube. This implies that if the section area decreases, velocity must increase for the mass flow rate to remain constant.

Law 2 tells us that the increase in velocity produces an increase in dynamic pressure. Now, the static pressure must reduce for the total pressure to remain constant. Based on this description of the venturi effect, the increase in dynamic pressure and decrease in static pressure must take place gradually as the section area reduces, not suddenly at the choke (trailing edge).

@AdityaSharma: there's a misunderstanding here. In a venturi tube the whole airflow flows inside it but in your picture "half" of the air is going over the plate and the other half under it: that's definitely not a venturi. Due to the blockage effect of the bottom wall I'd expect a lower speed on the belly i.e. a higher pressure and therefore lift.

@sophit yes, what is pictured seems more like a throttle (air inlet) plate on a carburetor.  That swirly area above and behind the plate trailing edge might be a good place to inject fuel.

@RobertDiGiovanni: yep, "butterfly valve" might be a good starting point to understand that airflow

@sophit you say that it cannot be considered a Venturi since not all of the flow is entering it. Admittedly, I do not understand how the flow would behave at the entrance of the "butterfly valve", but after half of the flow has entered the "duct" (formed by the "butterfly valve" and the wind tunnel walls), what would prevent the duct from behaving like a Venturi?

@AdityaSharma look at a stone in a river.  The momentum of incoming fluid creates a rise in pressure when any obstacle slows it.  Consider each half of a Venturi as an obstacle.

@RobertDiGiovanni But then what we observe in practice is that the static pressure decreases and the velocity increases through the Venturi. The same will happen around the stone. The velocity of water flow must be increased in order for a constant mass flow rate to be maintained at each section of the river. If flow doesn't speed up through a constriction, then that would violate the law of conservation of mass. You are right about that static pressure will be higher than free flow at the stagnation point, but then it will be lower than free flow around the stone.

@AdityaSharma then there is an alternative: mass flow is not constant.  Water can "back up" behind a dam.  Pressure can rise with air.  Interestingly,  keep the walls, but remove the two Venturi halves.  Now glue the long ends back together.  Looks like a airfoil.  Now put that back in the tunnel.

@RobertDiGiovanni "Then there is an alternative: mass flow is not constant" But then what about the law of conservation of mass? If the flow is backing up behind the dam, it's also backing up at the source, and at all points between the dam and the source. Even after the dam, the mass flow rate will be the same as that before the dam. If mass flow rate is altered at one point, it will be altered at all points within the stream (unless there is change in density resulting from change in pressure, an effect which is generally negligible in aerodynamics at speeds below Mach 0.3).

@AdityaSharma while the pressure is building the mass flow in the system may not be constant.  But once the pressure is high enough to force sufficient flow through the venturi then mass flow becomes constant:  flow in = flow out.  But we know (with our kitchen sink) if inflow is too great, and the drain too small, it overflows!  Is not a wind tunnel analysis where steady state is reached?  I think this is why a higher pressure stagnation area exists.

Let us continue this discussion in chat.

@AdityaSharma because the flow is subsonic: what happens in one point downstream depends on what happen upstream. If upstream half of the flow goes somewhere else then you have already lost the definition of Venturi and whatever happens downstream cannot be anymore related to a Venturi.

@sophit Then perhaps even Venturi tubes cannot be considered "Venturi", since some of the upstream flow enters them and some does not. Lets not call my apparatus a Venturi then, we'll call it a convergent duct instead. So how will the flow behave through this convergent duct, as far as it's static and dynamic pressures are concerned?

You need to consider more than just the static pressure. In the usual demonstrations of the Bernoulli effect, the pressure tap is placed perpendicular to the streamlines, which means it only "feels" the static pressure, but an object placed directly in the middle of the streamlines acts as a "stagnation point," and feels an additional pressure associated with actually forcing the streamlines to take a different path.
A small hole perpendicular to the streamlines can be used to measure the static pressure. A small hole placed parallel to the streamlines at a point where the streamlines are forced to stop can be used to measure the stagnation pressure. In this case, the airfoil is neither parallel nor perpendicular to the streamlines, and the resultant pressure will be somewhere in between the two.

Yes, but what will be the net effect? will the lift reduce due to change in pressure distribution? will it stay same? (or will it increase due to ground effect?)

@AdityaSharma The lower wall should increase the lift due to ground effect. The upper wall should decrease the lift. The net effect probably depends on the actual distances.

Alright, so if we remove the lower wall, the lift should reduce. By the way, I believe that the upper wall will also contribute towards "ground" effect, since it is also limiting the downwash, just like the lower wall (think of a Formula 1 car that uses ground effect to improve downforce - this is a similar case, just inverted). So if the F1 car with the same setup achieves an increase in lift (downforce), why would our setup suffer from a decrease in lift?

@AdityaSharma Hmm, I think you're right, actually. (The bit about the upper wall I was not so sure about to begin with). In that case, both the floor and the ceiling would increase the lift and it makes the overall effect much easier to figure out.

Yes, that is exactly what I thought when I first came across this problem, but then I thought: How would the apparent "reversal" of pressure distribution (as explained by Bernoulli's principle) affect the lift? - This is something I was never able to understand.


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Why do modern turbofans have swept blades?

(Picture sources: CF6 and GE90)

Aircraft have swept wings for two reasons:

It reduces the thickness-to-chord ratio of the effective chord, which is ideal for flight at higher Mach numbers.

At supersonic speeds, if the wing is swept behind the Mach cone, then the need for a sharp leading edge is eliminated, and a conventional round leading edge can be used instead - perfect for good takeoff/landing performance.

However, fan blades have an incredibly low t/c ratio as well as a sharp leading edge to begin with. So what's the point of having them swept?

I think it has indeed to do with the fact that the tips are supersonic (or with the fact that just before the tips the airflow is transonic). The design is proprietary so I don't know if there are papers or books around explaining that. Let's see if someone can give a decent answer :-)

Instead the proprietary, the designs can be patented.  See: https://patents.google.com/patent/US6071077A/en?oq=US6071077A  and https://patents.google.com/patent/EP2543818B1/en?oq=EP2543818B1  for example

@Adam Thank you for sharing that, didn't nearly read it cover to cover, but I can already tell it's really interesting!

Even helicopters have sweep tips now.

@user3528438 but helicopters have back swept wings. These are forward swept wings.

Lego uses swept blades for its helicopters too :-D

I think this question is a duplicate of https://aviation.stackexchange.com/questions/34798/why-do-some-turbofan-blade-have-forward-swept-tips?rq=1. The answer to that question and enclosed comment source state the answer: shock handling and surge protection. Although a good summary of that patent would be desirable...

@AdityaSharma as proposed by Adam in https://aviation.stackexchange.com/questions/34798/why-do-some-turbofan-blade-have-forward-swept-tips?rq=1#comment89243_34812, this patent gives a good (albeit long) overview: https://www.google.com/patents/US6071077

@U_flow That question you gave the link to is about forward sweep near the blade tips. I saw that question before writing this one - the answer there didn't answer my question. My question is: why the sweep is there in the first place? (and that over the entire length, not just tips). Also, my question might be answered in the patients (still haven't read them fully), but as such almost all the questions on ASE have been answered somewhere else; lets have an answer here as well :-)

I want to throw out that sometimes the answer is "just because."  Aerodynamics is the unholy spawn of observations and best available explanations.  "Me n' Ernie was working late one night at the wind tunnel and started farting around with nonsensically-curved blade ideas, and stumbled across an 8% performance gain."  The older I get the more I appreciate how flight is the mixing of observation with the best barely adequate explanations we have so far.  We were a hundred years into powered flight before we realized that our understanding of how a wing actually works was deeply flawed.

@AdityaSharma: maybe when you're finished with reading the patents you can write a proper answer.

See also this question Why do some turbofan blade have forward swept tips?, Although the answer given is a little vague and the question specifically around the tips. I will do my best to explain, but I am not a turbomachinery expert.
A compressor (or a fan) rotor stalls due to the breakdown of the tip leakage vortex or TLV (a bit like a wingtip vortex on an aircraft, caused as flow 'leaks' between the blade tips and the casing, spilling from the pressure to the suction side of the blade) at a certain blade, when this vortex breaks down its wake is incident upon the following blade, producing nasty recirculation zones, and stalling that blade. This leads to the development of stall cells, and potentially a compressor surge: the loss of steady mass flow through the rotor, (and therefore the thrust from the engine). Obviously, this is to be avoided. A modern turbofan engine however must operate over a wide range of flow velocities and densities, and so without the use of a variable geometry nozzle, keeping stall margins sufficiently wide is a huge challenge.
Because the blade tips in a fan are supersonic, there is a shockwave at the blade tip, which can exacerbate breakdown of the TLV, and so help induce a stall. It was discovered (see Bergner et al 2005) that sweeping the tips forward increased the distance between the shock and the TLV origin, decreasing their interaction and delaying breakdown of the vortex (hence, increasing stall margin). There are further effects to do with the boundary layer (forward sweep reduces the interaction between shock and BL), but that is well beyond my understanding.
In terms of back sweep nearer the hub on blades, this achieves the same ends as on transonic airliner wings: reducing the effective mach number of the incident flow, and delaying the formation of shockwaves on the fan blades, however, the downside is it adds a radial element to the flow velocity, and tends to push air outwards towards the tips, which could be detrimental to tip stall, so as always, everything is a balance.
The reason this has only been done recently, is simply: computing. High fidelity 3D CFD for rotors has only been around for a few decades, and so only now can turbomachinery engineers consider the aerodynamics of an entire blade, and even an entire rotor, rather than simply optimising airfoil sections in 2D. For a good overview of the various different factors around blade design this pdf is very helpful: https://rb.gy/kw5de
hope this is helpful.


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What was the cause of Qatar's A350 rejected takeoff on 2015-12 at JFK?
In December of 2015 Qatar Airways flew the first commercial flight of an Airbus A350 out of the United States, from JFK in New York City. The first attempt to take off was rejected, as documented at thepointsguy.com and in many articles derived from that one:

As you can see in the video above, about 18 seconds after we began rolling down JFK’s runway 22R, the aircraft self-aborted, bringing us from more than 100 mph to a loud, screeching halt in roughly 15 seconds.

The explanation given there isn't very detailed, and may not even be correct:

(Note that while Qatar representatives communicated that the plane aborted the takeoff on its own, it’s possible that the pilots may have reduced engine power manually, at which point the A350 automatically applied the brakes.)...
...Qatar’s Chief Commercial Officer, the highest ranking executive on the flight, came over to reassure us, explaining that an “indicator” was responsible for the abrupt abort. As I understand it, for some reason the A350 decided that our 11,000-foot runway was too short to support the takeoff, and the plane applied the brakes at full force — all on its own.

After about an hour and forty minutes they successfully took off from a different runway, and the flight was otherwise uneventful.
What actually happened here to cause the takeoff to be rejected? Was it aborted "automatically" as some articles have claimed or did the pilots take some action during the takeoff roll to abort it?
Obviously I'd prefer some documentation or good sources on what happened here, but it seems unlikely that there was ever a public investigation of this incident, so I'm also open to reasoning about what systems and/or procedures in the A350 (both Airbus and Qatar-specific systems and procedures of the time) are likely to cause this sort of thing, and (to help reduce and refute wild speculation), what systems likely did not cause this.

Seems to me that since the question is not asking for speculation about any accident, much less one currently under investigation, it is on topic.  Maybe someone will be able to provide a specific answer, or maybe not, we'll see.  Sometimes questions end up being unanswered for months or years and then someone answers.  And as noted, it appears that a comment has already provided content that would serve well as a partial answer.


text:
What size jet engines would it take to make a Cessna 140 hover?
I'm building a STOL bush plane using 4 jet engines located at 4 different points on the aircraft. I would like to know what size jet engines it would take to hover something like a Cessna 140.

You just need more thrust than it weighs. How much is it going to weigh? Maybe 2x the weight would do it.  And what happens if one of the engines flames out, or ingests a little bird or something? The engines will make a C140 a lot heavier, which means more thrust, which means heavier engines, which means more thrust needed, which means heavier engines, which means more fuel needed, which means more thrust needed, which means heavier engines, oops still more fuel needed. Eventually you end up with something that is very safe because it can't get off the ground.

Jet engine or rocket engine? Assuming the latter, do you have a supplier lined up?

Everyone’s been ignoring the most obvious problem with this. Jet exhaust at high temperatures, creates a forest or brush fire hazard associated with the operation of this aircraft.

@CarloFelicione although there have been multiple implementations of this concept: The F-35, the Harrier, numerous development constructs like the "Schwebegestell" https://en.wikipedia.org/wiki/VFW_SG_1262_Schwebegestell

Ah, but @U_flow your examples take off from improved, paved (or metal) runways. A "bush plane" (as the OP is planning on) will take off from unimproved grass/weed runways that, if not dry before the take off/landing attempt will rapidly become so at the exhaust temps of a jet engine, then quickly start burning.

@FreeMan The Harrier at least tried operating off grass and pierced metal planking - rather than fire the primary issue appears to have been more carving trenches into the surface that impacted safe landing and FOD issues to everything nearby. Though in the right (not damp UK?) conditions I'm sure something like this could first lightly toast the grass alight, then launch it afire across a wide area for optimal 'not get invited back' points.

Interesting, @GremlinWranger. Hadn't thought about it being an effective tool for bringing back trench warfare, but that does make sense, too.

The AV-8B has a maximum thrust of 23,500 lbf for a VTOL MTOW of 20,750 lbs.
For a plane the weight of the Cessna 140, which weighs 1450 lbs loaded, you'd need total thrust of about 1640 lbf. So 410 lbf each.
The excess thrust (TWR of 1.13) was needed to enable for a transition from hover to flight by rotating the nozzles. If you only use jets for hover, and retain a propeller and an engine for forward thrust, you'd be able to use a total thrust equal to total weight, a TWR of 1.
A modern plane that size can weigh less than the Cessna 140, as little as 1040 lbs. However, the engines will add 40 lbs each, not counting the need to carry two types of fuel if the main engine is an avgas piston, so you're at a minimum of 1200 lbs loaded.
The best new-build engine for your requirements is probably the PBS TJ150, although I can't be sure about cost and availability.
A less-advanced option is the AMT Lynx, which offers 350 lbf of thrust, but has worse fuel economy. It is meant for hobby and experimental designs though, and can be ordered for ~$55k per unit.
Overall, lift jets are an expensive way to get VTOL capability, but it's your project. Four engines will also burn 4-5 gallons per minute of hover, so account for that in the design. Rapid transition to horizontal flight and lift will minimize the fuel burn.
In general, jets are not a practical or safe option for a bush plane, since they produce heat and kick up ground debris, which can damage the engines themselves. It's safer and more fuel-efficient to spread power over more air, using propellers or rotors.

How many engines do you want to use?
In each VTOL, weight is key! As each engine you have to lift means, more duplicate weight (housing, fuel systems, starter etc.) it follows that you want as few engines as possible!!! This is also the reason why all jet-powered aircrafts have only one jet engine. In the following, I assume that you only install one jet. Last but not least, there simply are no jets (at least as far as I know) small enough such that you could install four of them. Either they weight more then your structure can handle or they produce far to much thrust.
Which engine do you need?
According to wikipedia, the gross weight of a Cessna 140 is about 658 Kg. Given that most VTOL aircraft have a thrust to weight ratio of about 1.1 to 1.3 you need a minimum of $1.1 \cdot 658 Kg = 724 Kg$ of thrust + the engine weight. Again, a quick look on wikipedia gives a nice table of engines from which I chose the PW530 with a weight of 280Kg and a thrust of around 12,84kN resulting in a lift to weight ratio of around $1284/(658+280)=1.37$ which should be enough also for the needed additional control system, fuel, cabeling etc.
Note that from that table, you see two things: This is one of the absolute smallest jet engine out there. There are other engines with less thrust (around 6kN) but they weight around 170Kg. Therefore when you had 4 of these, the engines would weight more then your aircraft. This means that a single engine is the weight-optimal (and cost optimal) way for such an aircraft.
Is it feasible to build a VTOL Cessna with Jet-engines?
No
Why?
Availability, cost, certification.

Making a VTOL Cessna 140 might be possible, but you'd probably end up needing to get a Supplemental Type Certificate if you intend to use the aircraft afterwards.
Firstly, the Continental O-190 engine that is usually installed on Cessna 140s runs on avgas, which means that you'd need to replace the engine with something like a CD-135 turbo-diesel engine if you intend to use one fuel tank for both the main engine and the lift jets. This engine does have the advantage of giving you 50 extra horsepower, but it also weighs 53 kilograms (115.4 lbs) more than the O-190.
Assuming that you find a CD-135 engine and install it, your Cessna 140 will have an empty weight of 457 kilograms (1,004.3 lbs). This rises to 533 kilograms (1,171.35 lbs) with 25 gallons of Jet-A fuel, since Jet-A is slightly heavier than Avgas. In order to achieve a TWR of 1 with 4 lift-jets, each lift-jet needs to produce roughly 295 lbf of thrust. In this case you'd probably need to use the PBS TJ150 turbojet, which weighs 37.7 lbs, and produces 337 lbf of thrust. Since you'll be using 4 of these engines, the Cessna 140 contraption will end up weighing 608.6 kilograms (1344.15 lbs). Conveniently enough, the 4 TJ150 engines will produce 1,348 lbf of thrust. This technically does mean that your Cessna 140 contraption will have a TWR above 1, but you wouldn't really be able to hover around in it since it would only take 4 pounds of additional weight to cause the TWR to fall below 1.
Secondly, even if you managed to shave off enough weight from the airframe so that you could actually get on the Cessna 140 contraption without causing the TWR to fall below 1, you'd need to find some way to control the aircraft while it's hovering, since control surfaces don't work while the aircraft isn't moving. Aircraft like the Harrier jump jet divert some of their engine's thrust to reaction control nozzles in order to provide pitch and roll authority during vertical flight but that wouldn't be possible in this case. You might be able to attach the turbojets to servo motors so that you could swivel them in order to get some degree of pitch and roll authority during vertical flight but this would only make the contraption even heavier.

Jet engines can run on avgas.  It makes them need maintenance more frequently, but they can do it.

True, turbines will run on literally any flammable liquid, but given that this contraption needs four of them and losing even one lift-jet would be enough to render the contraption unflyable, it's probably a good idea to just stick with what the manufacturer says the turbine is supposed to run on.

Controlling a quad-lift vehicle is actually very simple - a basic feedback from 6-axis gyro to a 2x2 matrix. That's how consumer drones work. Square-arrangement quadcopter is a very stable configuration.
The Harrier couldn't control nozzles one by one, it was manual, but still ran easily. The F-35B with its Y-arrangement would've been an instant crash without FBW.

Quadcopters are indeed very stable, as long as all four motors are working. Lose one motor and the quadcopter becomes a lawn mower.


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What would have been the pitch angle of this McCauley Prop before impact?
I'm trying to determine the pre-impact pitch of a two blade McCauley variable pitch on a Cessna 210 that hit shallow ocean water (8 feet, mostly sand) shortly after takeoff from an altitude of 300 feet at 140mph at a shallow angle of attack. I also have picture of the prop pitch knob, which seems to be pulled out for landing rather than pushed in for takeoff. I want to know if the pilot failed to gain altitude because the prop was never set for takeoff and it would leave a mark on the blade when it hit, showing what angle the blade was at. Any ideas based on the photos?

look up Ross R. Allen on Linkedin. He has dealt with the question of prop pitch effect on impact marks in airplane crashes in his consulting business. -NN

Any reason to believe that the FAA/NTSB won't be looking at this?

I'd consider looking for witness marks inside the hub due to other variable affecting how the blade bent.  But then the investigators would be trained in this.

The accident was already investigated and a report made by NTSB years ago. Pilot spatial disorientation and vestibular illusion. The plane was scrapped and I was able to obtain some of the parts as well as take pictures of the plane while it was at the plane graveyard. My ex was aboard that plane. NTSB did not mention the position of the pitch knob or the pitch of propellor at impact. Also, the altimeter read 2,400 feet but the only ascended to barely 400 feet. Pilot was instrument rated and experienced; but if his static port was blocked and his pitch mechanism was shot he had no chance.

I also have seen security camera videos of the predawn takeoff  and crash. He had a very shallow takeoff angle before turning right over water and then descended gradually from max 300 feet to zero feet. Elevator trim was found to be neutral so I don't think he was pitching his nose down because of the illusion of ascending while accelerating. He was an experienced pilot and instrument rated. But if his altimeter static port was blocked and frozen at 2,400 feet (probably forgot to check it when he landed), and he also either forgot to push in the prop pitch knob OR that also was shot...

A link to the report would be useful then. In my opinion, it's impossible to tell now. The knob in particular, given the amoun of time passed, and the amount of hands handling it, there's no guarantee that's in the position it was before the crash.

Upon my inspection, the knob was in a fixed position and corroded with salt water in that position. It had been under water for several hours and in storage for over a year. None of those items were inspected on the report. McCauley declined to help me by answering technical questions regarding how to take the internal hub assembly apart to determine pitch angle at impact. I've spent three years examining all evidence and the final report, which doesn't state which instruments the pilot might have been having trouble with. Also, the passenger turned iphone flashlight on seconds before impact

@Argonaut5 I'm so sorry for your loss. I hope whatever comes if this it will bring you peace. God bless you.

I'm sorry for your loss.
It's probably impossible to determine the pitch of the propeller at the time of impact just from the propeller damage. You may be able to confirm or reject your hypothesis based on the recorded performance of the aircraft, however.
From your description, this sounds like it's probably referring to NTSB accident number ERA19FA266. The ADS-B track in that report indicates that the plane reached its peak altitude (which the final report says was 425 feet indicated) around 20 seconds after takeoff, while the final report indicates that it accelerated from 70 knots at liftoff to about 100 knots at its highest point. (As an aside, the existence of ADS-B data indicates that the altimeter was operative at the time of the crash).
In other words, it was able to maintain an average climb rate of around 1200 fpm while in a bank and simultaneously accelerating 30 knots in 20 seconds. A Cessna 210B manual I found online indicates that the expected rate of climb at vy is between 1270 and 1900 fpm, so this doesn't sound like an airplane struggling to climb.
It's a sad fact that many fatal cases of spatial disorientation involve a pilot who is instrument rated and current. The pilot departed VFR (as they had not called ATC to pick up a clearance) and may have been depending on visual references. The sudden loss of visual references as he turned over a dark ocean on a cloudy early morning may have caused spatial disorientation that was simply too much to overcome.

Thank you so much Chris for your response and observations. Your comment on the altimeter was very helpful. I took a picture of the post accident altimeter and the dials read about 2,400 feet so I had thought the dials were stuck from a blocked static port  at that altitude prior to the crash. The passenger had turned her iphone flashlight  on a few seconds before impact, so perhaps he had trouble seeing the instruments. I also accept your assessment of the climb rate. It's still hard for me to grasp why he left so early when 15 minutes later the visibility was so much better. Thanks so much!


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What difference does altitude make on fuel efficiency?
Most commercial flights are between 29,000 and 42,000 feet in altitude apparently. But I realise I don't know why.  My guesses are:

If you fly too high the air pressure gets very low and you have to use more fuel flying faster to get the same lift...?
But why don't commercial flights fly at 10,000 feet, say, at least over the Atlantic where no one can complain about noise?

Optimum cruising altitude is a delicate balance between air density, lift, drag, and engine performance.
Thinner air at high altitudes means less drag, but also less lift and less power.
Today’s modern jetliners are optimized for about 35,000-40,000 feet for the best speed and fuel efficiency. Weather and wind direction can sometimes cause lower altitudes to be more efficient.

As usual in the aerospace world, it is a matter of compromises. First the pros of flying high and then the cons.

Commercial jetliners follow a flying path which resembles a stairs where the flying altitude is increased in steps till the limit of some 10km.
This increase in altitude is due to the fact that, during its flight, a jetliner diminishes its weight due to fuel's consumption. In order to compensate for the lower and lower weight, lift must be reduced accordingly and that can be achieved either decreasing the AoA (i.e. $C_l$) or decreasing the air density $\rho$:
$L=½ \rho V² S C_l$
Decreasing $\rho$ is easy since you just need to fly higher. Changing AoA would also be easy to obtain but the aerodynamic shape of a jetliner is optimised to be very efficient within a very narrow range of AoA; plus, try to push a cart along the aisle with a -5° attitude :)
Another positive effect of flying high is that, obeying the drag a very similar equation like the one of the lift, less air density implies less drag as well.

Flying high would appear so far to be a good thing, so why not flying higher?.
At higher altitudes engine's thrust reduces as well: said 1 the thrust generated by a modern turbofan as measured on the testbench, at 10km height and Mach 0.8 that thrust can be as less than 0.3. And here we hit a first limitation: going higher and higher thrust decreases more and more but faster than the decrease in drag and at a certain height there's simply no thrust left to counteract drag or to manoeuvre the aircraft. This height is called ceiling and for example for an A320 is some 12.5km. Obviously if there is a need to fly higher (Concorde for example), thrust can be increased but at the expense of a higher fuel consumption $\Rightarrow$ more fuel $\Rightarrow$ more weight.
Another limitation of flying higher is structural: the cabin has to be pressurised to a value which is comfortable and let the passengers freely breathe. So the fuselage is basically a tank and the weight of its structure is proportional to the pressurisation level. Flying higher would imply the need of a higher pressurisation and therefore a heavier fuselage with all that implies: more weight $\Rightarrow$ more lift $\Rightarrow$ more drag $\Rightarrow$ more thrust $\Rightarrow$ more fuel consumption $\Rightarrow$ more fuel $\Rightarrow$ more weight.
A third limitation is that in case of a sudden depressurisation, if the altitude were too high, passengers wouldn't even have the time to put the oxygen mask on their faces. That's the main reason why the Concorde had so small windows and a very powerful pressurisation system.
Another limitation is due to stall speed. We rewrite the previously written lift equation in $V$:
$V=\sqrt{\frac{L}{½ \rho S C_l}}$
Now, $L$ equals weight $W$ and $C_l$ is limited by its maximum value $C_{l_{max}}$ before stall (which is more or less 1.3 for a jetliner without flaps deployed). Substituting these values we get the minimum speed at which the airplane can still fly before stalling:
$V_{min}=\sqrt{\frac{W}{½ \rho S C_{l_{max}}}}$
As it can be seen, this minimum speed before stall increases at higher altitude due to the decrease in $\rho$. Flying higher would therefore imply the need of flying faster to stay well away from the stall condition. Obviously to fly faster an airplane needs more powerful engines with the same negative implications as before.

One bonus problem with flying too high: Lift also decreases with air pressure (or even squared to it? i dont know the details). So you'd need bigger wings to go higher, which produces more drag which needs to be countered by engines that loose efficiency and so on

I cross the USA a lot in an automobile and I watch my MPG.  Guess what happens to my MPG when I'm in high-altitude Utah-Wyoming.
Drag (which causes fuel consumption) has two components.
Induced drag
This is the additional "drag" created by using the wings to create lift to carry the weight of the plane.  The induced drag is proportional to vehicle weight.
My car does not have induced drag because it sits on wheels.  I do have some rolling resistance that is proportional to weight (because of the way rubber tires work), but trains don't (to speak of; it's inconsequential by comparison to any other mode).
This induced drag is not significantly affected by altitude.
Plain old aerodynamic, or parasitic drag
This is exactly the same aerodynamic drag that airframe shape would have if its weight was fully supported by railroad wheels.  Simply pushing that shape through the air at 600 mph takes a lot of energy.
This is very affected by altitude - like with my car getting better MPG at high altitudes.  Same with an airplane - the higher you fly the better your "MPG" as it were.

So induced drag is a wash, but parasitic drag improves as you climb.

All the other answers already mention that drag get lower

@sophit Of course they do since it's correct.  However, the art is in how you explain it, and those other answers explain it at different levels of complexity. I think breaking down the two types of drag is helpful, and no one else was covering it in an accessible way.

Ah ok, at poetry level you mean  Anyway induced drag is also proportional to density so it does reduce with altitude just like the "plain old aerodynamic drag": $D=½\rho V²S(C_{d_0}+C²_l/πARe)$

+1, but I'm not fully convinced by your car example: While drag is of course a factor, I would be very surprised if its effect on your gas mileage were not eclipsed by however your engine and ECU react to the lower oxygen level.

Another factor here could be temperature.  Higher altitudes are generally colder, and gas-powered cars tend to be slightly more fuel efficient in colder climes (unless it's so cold there's snow and ice on the road, as the lack of traction could reduce efficiency).  You're also not likely running your A/C in cold weather, which also decreases efficiency.  (The heater does not, however, since it just passively vents in heat off the engine.)

@sneftel My father worked in automotive emissions control so I kept a close eye on the field. Car emissions are astonishingly tight; combustion is dead-nuts stoichiometric at all conditions. My car gets smog checked and passes. They use mass airflow sensors (which detect altitude passively), barometric sensor (inside the ECU) and feedback with o2 sensor on the exhaust. I don't see any combustion efficiencies to gain at altitude.

@Darrel Good point about X-factors like that. I don't have A/C or power steering. At speeds over 40 mph I always run windows closed, blower at max. I've crossed the country in all seasons, 0F/-18C and 100F/37C, with lowland temps being fairly near high altitude temps.   MPG is always 20-30% better at altitude. But it's better on hot days. And yeah, worse in rain and snow because plowing water costs energy (the EV guys note the same).


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Should I advance my propeller lever forward during a glide?
I am reading the FAA Airplane Flying Handbook, chapter 3. It says:

To enter a glide, the pilot should close the throttle and, if equipped, advance the propeller lever forward. (...)

Why do we advance the propeller lever forward? Doing so will increase RPM but will also increase drag, right? Whereas in a glide, we should reduce drag to a minimum. So shouldn't we "feather" the propeller instead to reduce drag?

Isn’t that the same thing?

@Frog Propeller lever forward gives a fine blade pitch - the opposite of feather.

Don't think of it as resetting propeller pitch. Think of it as resetting the propeller governor to let the engine run at redline RPM when at max throttle. Outside of engine failure situations, glide ratio effected by propeller blade angle is not a consideration.
Going to wide open throttle with the propeller governor set to, say, 2200 RPM does a few things:

The power available is only that which WOT gives at 2200 RPM, maybe 85% of rated horsepower, when you may want all of it.
WOT at low RPM raises Brake Mean Effective Pressure in the cylinders to very high levels and can bring on detonation if the conditions are right.
Lastly if you go to WOT and then follow up with max propeller RPM applied suddenly, the propeller will typically overspeed past redline momentarily as the governor overshoots the RPM setting and has to catch up. Everybody for miles around can hear the result.

There are times when you may leave the prop RPM where it is, say if you are reducing power for an enroute descent, and when you get to the level off altitude, you're just going to bring manifold pressure gently back to its original cruise value. But for any descent where there is a likelihood of needing full power at the bottom of the descent, you ease the prop to max RPM after reducing throttle to ensure that full power is available as soon as throttle is advanced.

The answer here can be fairly nuanced. But this is a great example of a time when you may need to deviate from the advice in the AFH and follow your POH. Remember- always follow the POH when it disagrees with the AFH, PHAK, or basically any other document!
In a simulated engine out, it makes sense to leave the propeller at fine pitch for a couple reasons. First is that you may need to go around, and you don't want to apply full power when the propeller control is at low RPM. Second is that the idle engine is still producing some power. The pitch being as fine as possible can help offset some of this power and keep you from expecting too much of your glide ratio in a real emergency.
In a bona fide engine out, there are a few possibilities:

in some airplanes or with some failure modes you will lose pitch control over the propeller when the engine dies, so it's a moot point
if you can feather the propeller, you almost certainly should, as it significantly reduces drag
some airplanes do not have the ability to feather the propeller. Putting the propeller to high RPM may cause more drag. But it may also cause the propeller to stop windmilling and give a net reduction in drag! See what, if anything, your POH says to do with the propeller control in an engine out

Unfortunately the handbook does not give any explanation why they propose the contra intuitive adjustment.
I for myself see two possibilities:

As the handbook states, for a powered plane a glide with the engine stopped is an abnormal situation except the final approach before touchdown. A pilot should try to restart the engine by any chance. In case the propeller was feathered the force to rotate it by the crank might be to high to start the engine. Having the propeller lever forward eliminates this  problem.

For a powered plane with constant speed propeller the touchdown configuration is with the propeller lever forward so that, in case of a go around, the plane is in start configuration. Therefore a pilot is used to how the descend of the plane feels with the propeller set to maximum pitch (and drag). This might support an "on point landing" in unknown terrain better then having a slightly longer distance to travel.

This is a guess, but one possible answer is that at idle power, the engine is not putting out any torque, so it will be spinning at lowest possible RPM for the pitch angle set by the prop governor, so pushing the prop lever forward, (which flattens the pitch) will not cause as much of an increase in rpm (and concomitant increase in drag), as it decreases drag by flattening the pitch.
But this is a guess...


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What is the purpose of variable stator vanes in the power turbine part of a free shaft gas turbine?
both M1 Abrams tank and T-80 tanks have free shaft gas turbine engines and both of their engines have variable stator vanes in their power turbine part. It says in the picture below that these variable stators exist to prevent engine damage during gearshifts. How could variable stators prevent engine damage during shifts? By the way, PCA talked about in the picture is the automatic fuel injection system that regulates fuel injection during gear changes.

There is no clutch mechanism in the traditional sense in the transmission of the T-80 tank. Its role is played by the above-mentioned PCA. When changing gears, the PCA pedal is moved to a stop. In this case, the nozzle blades rotate 70-80° from the position corresponding to the maximum tractive power of the turbine, which actually leads to its stop (zero power position). Moreover, it is possible to move the voter lever only when the PCA pedal is depressed, since in the initial position it mechanically and electrically blocks the hydraulic servo gearshift. Thus, the guaranteed protection of the turbine rotor from misuse is ensured. By the way, the absence of such a mechanism in the GTE-ZT engine of the object 167-T led to the destruction of the turbine blades during the tests.

Also a picture of the AGT-1500 from M1 Abrams where it also shows the variable power turbine stator blades:

As asked, this question is unrelated to aviation. Please edit so that it has some cognizable tie to the subject of this exchange. "Gear shifts" are used with the transmission of battle tanks, but not as much with aircraft engines.

@RalphJ It isnt directly releated yeah but its still very much so releated to jet engines etc thats why i thought it would still be appropriate to post it. Because this is basically same engine construction as turboshaft engine in a helicopter.

@sophit partially yeah, but the answer below by John K provided all the answers i was looking for.

When you take a turboshaft engine and subject it to large and sudden load changes, that it doesn't have to cope with in an aircraft, like during gear ratio changes in a tank, you get very sudden reductions or increases in power turbine RPM as the torque demand fluctuates.
The airflow having a lot of inertia, sudden changes like that create large changes in local angle of attack (AOA) of the power turbine blades themselves. The blades become unloaded from a reduction in AOA, or conversely they become too loaded from a sudden increase in AOA and can aerodynamically stall.
These changes in airflow at the power turbine effects the upstream flow, disrupting it, and when you disrupt the airflow in the gas generator section (the compressor/burner/HP-LP turbines), you can get compressor stalling, leading to surging of the gas generator, leading to loss of control of the flame in the burner can (which is contained by creating a boundary of cooler air around the flame using vent holes), leading to too much heat leaving the burner can, leading to overtemps in the HP/LP/power turbines. And thus the engine gets damaged.
So the variable stator vanes are to compensate for sudden flow changes from sudden changes in torque load on the turbine caused by having to deliver power though a gearbox with different gear ratios, by redirecting the incoming flow to keep the angle of attack on the power turbine blades within an ideal range.
If you built a tank with a Continuously Variable Transmission, where you can set a constant engine RPM and only vary it gradually, you'd be closer to an aircraft turboshaft/turboprop environment and you could probably dispense with the variable stators for the power turbine.

I wonder if hydraulic drive would help.

The transmission is hydromechanical, there's a fancy name for it, and the engine is probably connected through a torque converter, but  it has 4 distinct fwd gear ratios and the key is being able to vary torque to the tracks while allowing the engine to wind up and down at its own pace. The power turbine stators probably give the Lycoming turboshaft in the M1 almost piston engine throttle response.

Its actually not hydromechanical because it does not have torque converter, it also does not have any sort of clutch mechanism thanks to free shaft gas turbine kinda acting like a fluid coupling and disconnecting engine from drivetrain at idle. It has 2 side planetary gearboxes with integerated final drives called BKP. They are manually shifted, Here is a picture https://media.discordapp.net/attachments/607630776812568595/1080525344437715044/transmission_t-80.png , https://media.discordapp.net/attachments/630092657972084746/1058704209706483742/scale_2400.png

Ahh great thanks. That's for the T-80? The MI has a transmission called a "hydrokinetic" transmission.

@JohnK Yeah the first picture is for T-80 and the second picture is for T-72 which has nearly identical transmission. Only difference is in T-80 BKP they deleted a planetary  gearset and used the leftover space to reinforce the brakes and clutches to handle the big torque of gas turbine. Otherwise their BKPs are same. M1 has hydromechanical because it also has torque converter. In russian literature any transmission coupled to a TC is referred as hydromechanical transmission. Maybe in English it is referred as hydrokinetic im not sure.

Also a picture of the M1 Abrams transmission X1100 , TC is visible as well https://media.discordapp.net/attachments/630092657972084746/974611148185956362/unknown.png?width=851&height=676

X1100 looks like a beefed up automatic transmission from a truck lol.

Let us continue this discussion in chat.

excellent answer, +1, thanks for sharing it here. -NN